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K-Similar Strings

Difficulty: Hard

Problem Description

Strings s1 and s2 are k-similar (for some non-negative integer k) if we can swap the positions of two letters in s1 exactly k times so that the resulting string equals s2. Given two anagrams s1 and s2, return the smallest k for which s1 and s2 are k-similar.

Key Insights

  • The problem involves transforming one string into another using a limited number of swaps.
  • Both strings are anagrams, meaning they contain the same characters in different orders.
  • The minimum number of swaps required can be determined by examining the positions of mismatched characters.
  • A breadth-first search (BFS) approach can efficiently explore the possible configurations of the string through swaps.

Space and Time Complexity

Time Complexity: O(n!), where n is the length of the strings (due to permutations in the worst case). Space Complexity: O(n), for the queue used in BFS and to store visited states.

Solution

The solution employs a breadth-first search (BFS) algorithm to explore all possible configurations of the string s1 by performing character swaps. The main idea is to use a queue to keep track of the current state of the string and the number of swaps made so far. At each step, we identify the first position where the characters of s1 and s2 differ, and we attempt to swap that character with other characters in the string to create new configurations. We continue this process until we find a configuration that matches s2, at which point we return the number of swaps made.

Code Solutions

from collections import deque

def kSimilarity(s1: str, s2: str) -> int:
    if s1 == s2:
        return 0

    queue = deque([(s1, 0)])  # (current string, number of swaps)
    visited = set([s1])

    while queue:
        current, swaps = queue.popleft()
        if current == s2:
            return swaps

        # Find the first position where characters differ
        i = 0
        while i < len(current) and current[i] == s2[i]:
            i += 1

        # Try all possible swaps
        for j in range(i + 1, len(current)):
            if current[j] == s2[i]:  # Only swap if it matches the target
                # Generate the new string by swapping
                next_state = list(current)
                next_state[i], next_state[j] = next_state[j], next_state[i]
                next_state = ''.join(next_state)

                if next_state not in visited:
                    visited.add(next_state)
                    queue.append((next_state, swaps + 1))
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